Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

int N, M, S, T;

int shortest_path(vector<vector<int>> G){
    vector<int> d(N, -1);
    
    queue<int> que;
    que.push(S);
    d[S] = 0;
    
    while(!que.empty()){
        auto u = que.front(); que.pop();
        for(auto v: G[u]){
            if(d[v] == -1){
                d[v] = d[u] + 1;
                if(v == T)
                    return d[v];
                que.push(v);
            }
        }
    }
    return -1;
}


vector<ll> bfs(vector<vector<int>> G, int s){
    vector<int> d(N, -1);
    
    queue<int> que;
    que.push(s);
    d[s] = 0;
    
    while(!que.empty()){
        auto u = que.front(); que.pop();
        for(auto v: G[u]){
            if(d[v] == -1){
                d[v] = d[u] + 1;
                que.push(v);
            }
        }
    }
    
    vector<ll> res(N);
    for(int i=0; i<N; i++) if(d[i] >= 0)
        res[d[i]]++;
    return res;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    cin >> N >> M >> S >> T; S--; T--;
    
    vector<vector<int>> G(N);
    for(int i=0; i<M; i++){
        int x, y; cin >> x >> y; x--; y--;
        G[x].push_back(y);
        G[y].push_back(x);
    }
    
    int sp = shortest_path(G);
    auto st = bfs(G, S);
    auto tt = bfs(G, T);
    
    ll ans = 0;
    for(int i=0; i<=sp-2; i++)
        ans += st[i] * tt[sp-2-i];
    
    cout << ans << endl;
    return 0;
}